Question

Prove that
${|z_1+z_2|}^2+{|z_1-z_2|}^2={2\left|z_1\right|}^2+2{\left|z_2\right|}^2$.
Interpret the result geometrically and deduce that
$|\alpha +\sqrt{{\alpha }^2{-\beta }^2}|+|\alpha -\sqrt{{\alpha }^2-{\beta }^2}|=|\alpha +\beta |+|\alpha -\beta |$,
All numbers involved being complex.

Answer 1

${|z_1+z_2|}^2+{|z_1-z_2|}^2$
$=(z_1+z_2)\left(\overline{z_1+z_2}\right)+(z_1-z_2)\left(\overline{z_1-z_2}\right)$
$=(z_1+z_2)(\overline{z_1}+\overline{z_2})+(z_1-z_2)(\overline{z_1}-\overline{z_2})$
$2z_1\overline{z_1}+2z_2\overline{z_2}$ [other terms cancel]
$={\left|z_1\right|}^2+2{\left|z_2\right|}^2$
Geometrical Interpretation.
Let $P$ and $Q$ be the points of affix of $z_1$ and $z_2$ respectively.
Complete the parallelogram $OPRQ$. Then $R$ represents the point $z_1+z_2$.
Hence $\left|z_1\right|=OP$, $\left|z_2\right|=OQ$, $|z_1+z_2|=OR$ and $|z_1-z_2|=QP.$
We know that the sum of squares of the sides of a parallelogram is equal to the sum of squares of its diagonals, that is
$2O{P}^2+2O{Q}^2=O{R}^2+Q{P}^2$
or $2{\left|z_1\right|}^2+2{\left|z_2\right|}^2={|z_1+z_2|}^2+{|z_1-z_2|}^2$
Deduction : Let $z_1=\alpha -\sqrt{{\alpha }^2-{\beta }^2}$
and $z_2=\alpha -\sqrt{{\alpha }^2-{\beta }^2}$.
$\therefore z_1+z_2=2\alpha$,
$z_1-z_2=2\sqrt{{\alpha }^2-{\beta }^2}$ ...$\left(2\right)$
$z_1{z}_2={\alpha }^2-({\alpha }^2-{\beta }^2)={\beta }^2$ ...$\left(3\right)$
Square both sides of the result to be proved.
${\left|z_1\right|}^2+{\left|z_2\right|}^2+2\left|z_1{z}_2\right|={[|\alpha +\beta |+|\alpha -\beta |]}^2$
$L.H.S.=\frac{1}{2}[|z_1+z_2|+{|z_1-z_2|}^2]+2\left|z_1{z}_2\right|$
$=\frac{1}{2}[{\left|2\alpha \right|}^2+2{\left|\overline{{\alpha }^2-{\beta }^2}\right|}^2+2\left|{\beta }^2\right|]$
$=[2{\left|\alpha \right|}^2+2{\left|\beta \right|}^2+2|\alpha +\beta ||\alpha -\beta |]$
$={|\alpha +\beta |}^2+{|\alpha -\beta |}^2+2|\alpha +\beta ||\alpha -\beta |$
$={[|\alpha +\beta |+|\alpha +\beta |]}^2$
Proceed as in last part.